How to calculate basis of kernel?
Sophia Terry 
I have a linear transformation.
The transformation and what I tried is written on the attached work page.
Is my way wrong? what is the basis of KerT?
LinearAlgebra: S -> S is a joke with my friends. sorry for this.
1 Answer
$\begingroup$Your way isn't wrong, per se, but it could be a great deal more efficient. The more calculations you go through, the more chances you have to make a mistake. Indeed, you have made an error. Namely, when you were at $$\left[\begin{array}{ccc}0 & 1 & 2\\-1 & 1 & 3\\-1 & 0 & 1\end{array}\right]$$ and took $R_3-R_2,$ you should have gotten $$\left[\begin{array}{ccc}0 & 1 & 2\\-1 & 1 & 3\\0 & -1 & -2\end{array}\right],$$ rather than $$\left[\begin{array}{ccc}0 & 1 & 2\\-1 & 1 & 3\\0 & -1 & 2\end{array}\right].$$ From there, you did everything else right, but that mistake yielded an erroneous result. You've also misinterpreted this result. If you were able to row reduce to a matrix such as you have in the bottom-right, that would, in fact, mean, that for $T(x,y,z)$ to be null, then we need $x=0,y=0,z=0$. That is, your kernel is simply the origin, the subspace of dimension $0$. This has the empty set as a basis, not the set $\bigl\{(-1,1,1)\bigr\}.$
To be more efficient, I'd start with $R_2-2R_1$ and $R_3-3R_1$, yielding $$\left[\begin{array}{ccc|c}1 & 2 & 3 & 0\\0 & -1 & -2 & 0\\0 & -2 & -4 & 0\end{array}\right].$$ In this way we have all zeroes below the top left entry (our first non-zero leading entry). Next, I'd take $R_3-2R_2$, yielding $$\left[\begin{array}{ccc|c}1 & 2 & 3 & 0\\0 & -1 & -2 & 0\\0 & 0 & 0 & 0\end{array}\right],$$ thus giving us all zeroes below the $-1$ in the second row (our second non-zero leading entry). Then, I'd take $-1\cdot R_2$ (so now all of our non-zero leading entries are $1$s), and finally $R_1-2R_2$, yielding $$\left[\begin{array}{ccc|c}1 & 0 & -1 & 0\\0 & 1 & 2 & 0\\0 & 0 & 0 & 0\end{array}\right].$$ Our non-zero leading entries, now, are all $1$s, and have all $0$s above and below them. This is called the reduced row echelon form of our original matrix.
Rewriting the three rows of this augmented matrix as their equivalent equations, we have: $$x-z=0$$ $$y+2z=0$$ $$0=0.$$ That last is trivial, so we have that $(x,y,z)$ is in the kernel of $T$ if and only if $x=z$ and $y=-2z$. In other words, the kernel of $T$ consists of all vectors of the form $(z,-2z,z)$ with $z$ real, or equivalently, all real scalar multiples of the vector $(1,-2,1)$. Hence, $$\bigl\{(1,-2,1)\bigr\}$$ is a basis for the kernel of $T$.
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